Class 12 Physics Top 10 Must-Know 5-Mark Derivations with Proofs (CBSE & BSEB 2026/2027) | SolvIQ PrepOne
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Class 12 Physics Top 10 Must-Know 5-Mark Derivations with Proofs (CBSE & BSEB 2026/2027)

PrepOne Academic Team
September 18, 2026
15 min read
Class 12 Physics Top 10 Must-Know 5-Mark Derivations with Proofs (CBSE & BSEB 2026/2027)

Summary: This master revision guide covers the Top 10 most repeated 5-mark derivations in Class 12 Physics for CBSE and State Boards (BSEB). Each derivation includes the statement, diagram blueprint, step-by-step mathematical proof, and examiner step-marking distribution.

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In both CBSE (Section E, 3 questions ×\times 5 marks = 15 marks) and Bihar Board (Section B, 3 questions ×\times 5 marks = 15 marks), long-answer questions are almost exclusively drawn from a pool of foundational derivations. Mastering these 10 derivations secures full marks on the theoretical component of your physics board paper.

Top 10 Physics Derivations Directory

1. Electric Field Due to an Infinitely Long Straight Wire (Gauss's Law)

Unit I: Electrostatics

Consider an infinitely long thin wire having uniform linear charge density λ\lambda. By cylindrical symmetry, choose a coaxial Gaussian cylinder of radius rr and length ll.

EdA=curvedEdAcos0+endsEdAcos90=E(2πrl)\oint \vec{E} \cdot d\vec{A} = \int_{\text{curved}} E \, dA \cos 0^\circ + \int_{\text{ends}} E \, dA \cos 90^\circ = E(2\pi r l)
qenclosed=λlq_{\text{enclosed}} = \lambda l
By Gauss’s Law: E(2πrl)=λlε0    E=λ2πε0r\text{By Gauss's Law: } E(2\pi r l) = \frac{\lambda l}{\varepsilon_0} \implies E = \frac{\lambda}{2\pi \varepsilon_0 r}

Examiner Tip: State explicitly that the flux through the circular flat caps is zero because EdA\vec{E} \perp d\vec{A} (cos90=0\cos 90^\circ = 0).

2. Electric Field Due to a Uniformly Charged Thin Spherical Shell

Unit I: Electrostatics

Let a thin spherical shell of radius RR carry total charge Q=4πR2σQ = 4\pi R^2 \sigma. Construct a concentric spherical Gaussian surface of radius rr.

Case 1: Outside Shell (rR):EdA=E(4πr2)=Qε0    E=14πε0Qr2=σR2ε0r2\text{Case 1: Outside Shell }(r \ge R): \quad \oint E \, dA = E(4\pi r^2) = \frac{Q}{\varepsilon_0} \implies E = \frac{1}{4\pi \varepsilon_0} \frac{Q}{r^2} = \frac{\sigma R^2}{\varepsilon_0 r^2}
Case 2: Inside Shell (r<R):qenclosed=0    E(4πr2)=0    E=0\text{Case 2: Inside Shell }(r < R): \quad q_{\text{enclosed}} = 0 \implies E(4\pi r^2) = 0 \implies E = 0

3. Parallel Plate Capacitor with Dielectric Slab of Thickness t<dt < d

Unit I: Electrostatics

Plates of area AA separated by distance dd. In vacuum thickness (dt)(d-t), electric field is E0=σε0E_0 = \frac{\sigma}{\varepsilon_0}. Inside dielectric of thickness tt, field is E=E0KE = \frac{E_0}{K}.

V=E0(dt)+Et=E0(dt)+E0Kt=E0[(dt)+tK]=Qε0A[dt(11K)]V = E_0(d - t) + E \cdot t = E_0(d - t) + \frac{E_0}{K} t = E_0 \left[(d - t) + \frac{t}{K}\right] = \frac{Q}{\varepsilon_0 A} \left[d - t \left(1 - \frac{1}{K}\right)\right]
C=QV=ε0Adt(11/K)C = \frac{Q}{V} = \frac{\varepsilon_0 A}{d - t(1 - 1/K)}

4. Drift Velocity and Microscopic Deduction of Ohm's Law

Unit II: Current Electricity

Free electrons under electric field EE experience acceleration a=eE/ma = -eE/m. With mean relaxation time τ\tau, drift velocity is vd=eEτmv_d = \frac{eE\tau}{m}.

I=neAvd=neA(eEτm)=(ne2τm)AE=σAEI = n e A v_d = n e A \left(\frac{e E \tau}{m}\right) = \left(\frac{n e^2 \tau}{m}\right) A E = \sigma A E
Since E=Vl,I=(ne2τAml)V    V=(mne2τlA)I=RI\text{Since } E = \frac{V}{l}, \quad I = \left(\frac{n e^2 \tau A}{m l}\right) V \implies V = \left(\frac{m}{n e^2 \tau} \frac{l}{A}\right) I = R I
where Resistivity ρ=mne2τ\text{where Resistivity } \rho = \frac{m}{n e^2 \tau}

5. Magnetic Field on the Axis of a Circular Current Loop (Biot-Savart Law)

Unit III: Magnetism

Radius RR, current II, point PP at axial distance xx. Resolving dBd\vec{B} into components; perpendicular components cancel by symmetry:

B=dBx=dBcosϕ=μ0I4π(R2+x2)RR2+x2dlB = \oint dB_x = \oint dB \cos \phi = \frac{\mu_0 I}{4\pi (R^2+x^2)} \frac{R}{\sqrt{R^2+x^2}} \oint dl
B=μ0IR4π(R2+x2)3/2(2πR)=μ0IR22(R2+x2)3/2B = \frac{\mu_0 I R}{4\pi (R^2+x^2)^{3/2}} (2\pi R) = \frac{\mu_0 I R^2}{2(R^2+x^2)^{3/2}}

6. Magnetic Field Inside a Long Solenoid (Ampère's Law)

Unit III: Magnetism

Consider a rectangular Ampèrian loop abcdabcd of length LL. The magnetic field outside is negligibly weak (B0B \approx 0), and along perpendicular segments bcbc and dada, Bdl\vec{B} \perp d\vec{l}.

Bdl=abBdl=BL\oint \vec{B} \cdot d\vec{l} = \int_a^b B \, dl = B L
Ienclosed=nLI(n=turns per unit length)I_{\text{enclosed}} = n L I \quad (n = \text{turns per unit length})
BL=μ0(nLI)    B=μ0nIB L = \mu_0 (n L I) \implies B = \mu_0 n I

7. Mutual Inductance of Two Long Coaxial Solenoids

Unit IV: Electromagnetic Induction

Two coaxial solenoids S1S_1 (radius r1r_1, turns n1n_1) and S2S_2 (radius r2>r1r_2 > r_1, turns n2n_2). When current I2I_2 flows in S2S_2, field inside is B2=μ0n2I2B_2 = \mu_0 n_2 I_2.

Φ12=N1(B2A1)=(n1l)(μ0n2I2)(πr12)=μ0n1n2πr12lI2\Phi_{12} = N_1 (B_2 A_1) = (n_1 l) (\mu_0 n_2 I_2) (\pi r_1^2) = \mu_0 n_1 n_2 \pi r_1^2 l \, I_2
M12=Φ12I2=μ0n1n2πr12lM_{12} = \frac{\Phi_{12}}{I_2} = \mu_0 n_1 n_2 \pi r_1^2 l

8. Refraction at Spherical Surface & Lens Maker's Formula

Unit VI: Optics

For single convex spherical refracting surface: μ2vμ1u=μ2μ1R\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}. Applying this across two surfaces of a thin lens in air (μ1=1,μ2=μ\mu_1 = 1, \mu_2 = \mu):

Surface 1: μv11u=μ1R1,Surface 2: 1vμv1=1μR2=μ1R2\text{Surface 1: } \frac{\mu}{v_1} - \frac{1}{u} = \frac{\mu - 1}{R_1}, \quad \text{Surface 2: } \frac{1}{v} - \frac{\mu}{v_1} = \frac{1 - \mu}{R_2} = -\frac{\mu - 1}{R_2}
Adding both: 1v1u=(μ1)(1R11R2)\text{Adding both: } \frac{1}{v} - \frac{1}{u} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)
By Thin Lens Formula (1v1u=1f):1f=(μ1)(1R11R2)\text{By Thin Lens Formula } \left( \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \right): \quad \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)

9. Proof of Laws of Refraction Using Huygens' Wavefront Theory

Unit VI: Wave Optics

A plane wavefront ABAB is incident at angle ii on interface separating medium 1 (speed v1v_1) from medium 2 (speed v2v_2). Time taken for secondary wavelet to travel from BB to CC is τ=BC/v1\tau = BC / v_1.

sini=BCAC=v1τAC,sinr=AEAC=v2τAC\sin i = \frac{BC}{AC} = \frac{v_1 \tau}{AC}, \quad \sin r = \frac{AE}{AC} = \frac{v_2 \tau}{AC}
sinisinr=v1v2=c/v2c/v1=μ2μ1=μ(Snell’s Law Verified)\frac{\sin i}{\sin r} = \frac{v_1}{v_2} = \frac{c/v_2}{c/v_1} = \frac{\mu_2}{\mu_1} = \mu \quad (\text{Snell's Law Verified})

10. Magnifying Power of Astronomical Telescope (Normal & Near Point Adjustment)

Unit VI: Optics

An astronomical telescope consists of an objective (fof_o, large aperture) and an eyepiece (fef_e, small aperture).

Normal Adjustment (Image at ):m=fofe,L=fo+fe\text{Normal Adjustment (Image at } \infty): \quad m = -\frac{f_o}{f_e}, \quad L = f_o + f_e
Near Point Adjustment (Image at D=25 cm):m=fofe(1+feD),L=fo+ue\text{Near Point Adjustment (Image at } D = 25\text{ cm}): \quad m = -\frac{f_o}{f_e} \left( 1 + \frac{f_e}{D} \right), \quad L = f_o + u_e

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